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\title{\heiti\zihao{2} 复变函数-第5章}
\author{20373963-樊若宸}
\date{\today}

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\maketitle
\section{求下列函数在其孤立奇点 (包括无穷远点) 处的留数.}

(1) $\dfrac{z}{(z-1)(z-2)^{2}}$;

\textbf{解:}\quad
函数有两个孤立奇点和一个无穷远点处的奇点.

$Res(f,1) = \lim\limits_{z\rightarrow 1}(z-1)\cdot\dfrac{z}{(z-1)(z-2)^2}=1$

$Res(f,2) = \left.\dfrac{1}{(2-1)!}\dfrac{\mathrm{d}}{\mathrm{d}z^{2-1}}[(z-2)^2\dfrac{z}{(z-1)(z-2)^2}]\right|_{z=2}=-1$

从而$Res(f,\infty) = - Res(f,1) - Res(f,2) = 0$

(2) $\dfrac{z}{(z-a)^{m}(z-b)}(a \neq b)$;

\textbf{解:}\quad
函数有两个孤立奇点和一个无穷远点处的奇点.

$Res(f,b) = \lim\limits_{z\rightarrow b}(z-b)\cdot \dfrac{z}{(z-a)^{m}(z-b)}=\dfrac{b}{(b-a)^{m}}$

$Res(f,a) = \left.\dfrac{1}{(m-1)!}\dfrac{\mathrm{d}^{m-1}}{\mathrm{d}z^{m-1}}[(z-a)^m\cdot\dfrac{z}{(z-a)^{m}(z-b)}]\right|_{z=a}=\dfrac{(-1)^{m-1}b}{(a-b)^{m}}=-\dfrac{b}{(b-a)^{m}}$

从而$Res(f,\infty) = -\left(Res(f,a)+Res(f,b)\right)=0$

(3) $\mathrm{e}^{\dfrac{1}{1-z}}$;

\textbf{解:}\quad
$z=1$是该函数的本性奇点.
$$
\begin{aligned}
    \mathrm{e}^{1/(1-z)} &= 1-\dfrac{1}{z-1}+\dfrac{1}{(z-1)^2}+\cdots
\end{aligned}
$$
从而$Res(f,1) = -1$, $Res(f,\infty)=1$

(4) $\sin \dfrac{1}{z}$;

\textbf{解:}\quad
$z=0$是该函数的本性奇点.

$Res(f,0) = c_{-1}=1$,从而$Res(f,\infty) = -1$.



(5) $\dfrac{z^{2 n}}{(z+1)^{n}}$ ( $n$ 为正整数);

\textbf{解:}\quad
$z=1$是函数的$n$阶极点.
$$
\begin{aligned}
    Res(f,-1) &= \left.\dfrac{1}{(n-1)!}\dfrac{\mathrm{d}^{n-1}}{\mathrm{d}z^{n-1}}z^{2n}\right|_{z=-1}\\
    &=(-1)^{n+1}C_{2n}^{n-1}
\end{aligned}
$$
从而无限远点的留数为:$Res(f,\infty) = -Res(f,-1)=(-1)^{n}C_{2n}^{n-1}$


(6) $\dfrac{1}{\left(\mathrm{e}^{z}-1\right)^{2}}$.

\textbf{解:}\quad
其存在无穷个奇点:$2k\pi i$.
$$
\begin{aligned}
    \dfrac{1}{\left(\mathrm{e}^{z}-1\right)^{2}} &= \left(\dfrac{1}{\mathrm{e}^{2k\pi i + (z-2k\pi i)}-1}\right)^2\\
    &=\left(\dfrac{1}{\mathrm{e}^{(z-2k\pi i)}-1}\right)\\
    &=\dfrac{1}{(z-2k\pi i)^2}-\dfrac{1}{z-2k\pi i}+P(z)
\end{aligned}
$$
其中$P(z)$是关于$z$在$2k\pi i$处洛朗展开的正则部分.从上式可见,对于每个孤立奇点,在其上的留数都为$-1$.


\section{利用留数定理计算下列积分, 其中 $\mathrm{C}$ 为正向圆周.}

(1) $\int_{C} \dfrac{1}{z \sin z} \mathrm{~d} z$, 其中 $C:|z|=1$;

\textbf{解:}\quad
考虑在$r<1$中,只有$z=0$是该函数的二阶极点.所以有:
$$
\begin{aligned}
    \int_{C} \dfrac{1}{z \sin z} \mathrm{~d} z&=2\pi iRes(f,0)
\end{aligned}
$$
所以
$$
\begin{aligned}
    \dfrac{1}{z\sin z}&=\dfrac{1}{z^2(1-\dfrac{z^2}{3!}+\dfrac{z^4}{5!}+\cdots)}\\
    &=\dfrac{1}{z^2}+\dfrac{1}{6}-\dfrac{z^2}{5!}+\cdots
\end{aligned}
$$
从而$\int_{C} \dfrac{1}{z \sin z} \mathrm{~d} z = 2\pi iRes(f,0)=0$

(2) $\int_{C} z \mathrm{e}^{\dfrac{1}{z}} \mathrm{~d} z$, 其中 $C:|z|=1$;

\textbf{解:}\quad
唯一奇点$z=0$.
$$
\begin{aligned}
    z\mathrm{e}^{1/z} &= z\left(1+\dfrac{1}{z}+\dfrac{1}{2z^2}+\cdots\right)\\
    &=z + 1 + \dfrac{1}{2z}+\cdots
\end{aligned}
$$
$\int_{C} z \mathrm{e}^{\dfrac{1}{z}} \mathrm{~d} z = 2\pi iRes(f,0)=2\pi \cdot\dfrac{1}{2} i = \pi i$

(3) $\int_{C} \dfrac{z}{(z-1)^{2}\left(z^{2}+1\right)} \mathrm{d} z$, 其中 $C:|z-1|=\sqrt{3}$;

\textbf{解:}\quad
由于所有奇点($1,\pm i$)都在$C:|z-1|=\sqrt{3}$内部,从而考虑无穷远点的留数.将函数在无穷远点的邻域展开:
$$
\begin{aligned}
    \dfrac{z}{(z-1)^{2}\left(z^{2}+1\right)} &= \dfrac{1}{z^3}\dfrac{1}{\left(1-\dfrac{1}{z}\right)^2\left(1-\dfrac{1}{z^2}\right)}\\
    &=\dfrac{1}{z^3}\left(1+\dfrac{2}{z}+\cdots\right)\left(1+\dfrac{1}{z^2}+\cdots\right)
\end{aligned}
$$
从而无限远点的留数为$0$.所以
$$
\int_{C} \dfrac{z}{(z-1)^{2}\left(z^{2}+1\right)} \mathrm{d} z = -2\pi i Res(f, \infty) = 0
$$



(4) $\int_{C} \dfrac{\mathrm{d} z}{(z-a)^{n}(z-b)^{n}}(n$ 为正整数, 且 $|a|<|b|,|a| \neq 1,|b| \neq 1)$, 其中 $C:|z|=1 ;$

\textbf{解:}\quad
当$1<|a|$时,函数在$C$的内部解析,从而$\int_{C} \dfrac{\mathrm{d} z}{(z-a)^{n}(z-b)^{n}}=0$

当$|a|< 1 < |b|$时,有
$$
\begin{aligned}
    Res(f,a)&=\dfrac{1}{(n-1)!}\left.\dfrac{\mathrm{d}^{n-1}}{\mathrm{d}z^{n-1}}[\dfrac{1}{(z-b)^{n}}]\right|_{z=a}\\
    &=(-1)^{n-1}C_{2n-2}^{n-1}\dfrac{1}{(a-b)^{2n-1}}
\end{aligned}
$$
从而
$$
\int_{C} \dfrac{\mathrm{d} z}{(z-a)^{n}(z-b)^{n}} = (-1)^{n-1}C_{2n-2}^{n-1}\dfrac{2\pi i}{(a-b)^{2n-1}}
$$

当$|a|< |b| < 1$时,有
$$
\begin{aligned}
    Res(f,b) &= (-1)^{n}C_{2n-2}^{n-1}\dfrac{1}{(a-b)^{2n-1}}
\end{aligned}
$$
由于$Res(f,a)+Res(f,b)=0$,故
$$
\int_{C} \dfrac{\mathrm{d} z}{(z-a)^{n}(z-b)^{n}} =0
$$



(5) $\int_{C} \dfrac{\mathrm{e}^{\mathrm{sin} z}}{z^{2}\left(z^{2}+1\right)} \mathrm{d} z$, 其中 $C:|z|=2$;

\textbf{解:}\quad
函数存在三个奇点:$0,\pm i$.
$$
\begin{aligned}
    Res(f,0)&=\left.\dfrac{\mathrm{d}}{\mathrm{d}z}[\dfrac{\mathrm{e}^{\sin z}}{z^2+1}]\right|_{z=0}\\
    &=1
\end{aligned}
$$

$$
\begin{aligned}
    Res(f,i)&=\lim\limits_{z\rightarrow i}\dfrac{\mathrm{e}^{\sin z}}{z^{2}(z+i)}\\
    &=\dfrac{\mathrm{e}^{\sin i}}{-2i}\\
    Res(f,-i)&=\lim\limits_{z\rightarrow -i}\dfrac{\mathrm{e}^{\sin z}}{z^{2}(z-i)}\\
    &=\dfrac{\mathrm{e}^{-\sin i}}{2i}
\end{aligned}
$$

从而
$$
Res(f,i)+Res(f,-i) = -\dfrac{\mathrm{e}^{\sin i}-\mathrm{e}^{-\sin i}}{2i} = \sin (i \sin i) =\sin\dfrac{\dfrac{1}{\mathrm{e}}-\mathrm{e}}{2}
$$


所以
$$
\int_{C} \dfrac{\mathrm{e}^{\mathrm{sin} z}}{z^{2}\left(z^{2}+1\right)} \mathrm{d} z = 2 \pi i\left(1+\sin\dfrac{\dfrac{1}{\mathrm{e}}-\mathrm{e}}{2}\right)
$$

(6) $\int_{C} \tan z \mathrm{~d} z$, 其中 $C:|z|=3$;

\textbf{解:}\quad
函数的奇点为$z = \pm \dfrac{\pi }{2}$,其都在$C:|z|=3$内部,从而考虑无穷远点的留数.
$$
\begin{aligned}
    \tan z &= \dfrac{\sin z}{\cos z}\\
    &=(z - \dfrac{z^3}{3!}+\cdots)\cdot\dfrac{1}{1-\dfrac{z^2}{2!}+\cdots}\\
    &=-\dfrac{2}{z^2}\left(z - \dfrac{z^3}{3!}+\cdots\right)\cdot\left(1+\dfrac{2}{z^2}+\cdots\right)
\end{aligned}
$$
从而$Res(f,\infty) = -2$,从而
$$
\int_{C} \tan z \mathrm{~d} z = -2\pi i Res(f,\infty) = 4\pi i
$$


(7) $\int_{C} \dfrac{1-\cos z}{z^{m}} \mathrm{~d} z(m$ 为整数 $)$, 其中 $C:|z|=\dfrac{3}{2}$.

\textbf{解:}\quad
当$m\leqslant 0$的时候,函数解析,从而$\int_{C} \dfrac{1-\cos z}{z^{m}} \mathrm{~d} z=0$

当$m=1$时,$Res(f,0)=0$.

当$m\geqslant 2$时,有
$$
Res(f,0)=\dfrac{1}{(m-1)!}\left.\dfrac{\mathrm{d}^{m-1}}{\mathrm{d}z^{m-1}}[1-\cos z]\right|_{z=0} = -\dfrac{1}{(m-1)!}\cos (\dfrac{(m-1)\pi i}{2})
$$

从而
$$
\int_{C} \dfrac{1-\cos z}{z^{m}} \mathrm{~d} z = -\dfrac{2\pi i}{(m-1)!}\cos (\dfrac{(m-1)\pi i}{2})
$$
\end{document}